The value of limn→∞ 13+23+…+n3n4 , is

A
$\frac{1}{2}$
B
$1$
C
$\frac{1}{4}$
D
$\frac{1}{8}$

Solution:

We have,

$lim_{n \to \infty} \frac{1^{3} + 2^{3} + \dots + n^{3}}{n^{4}} = lim_{n \to \infty} \frac{{\{\frac{n (n + 1)}{2}\}}^{2}}{n^{4}} = lim_{n \to \infty} \frac{(n + 1)^{2}}{4 n^{2}} = \frac{1}{4}$

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